Problem 50 - A Graphical Approach to Precalculus with Limits

Question

Write each vector in the form ai \(+\) bj. Round a and b to the nearest hundredth, if necessary. Direction angle \(210^{\circ},\) magnitude 3

Step-by-Step Solution

Answer

The vector is \(-2.60i - 1.50j\).

Step 1: Understanding the Problem

We are given a vector with a magnitude of 3 and a direction angle of \(210^{\circ}\). Our goal is to express this vector in terms of its horizontal \((ai)\) and vertical \((bj)\) components.

Step 2: Calculating the Horizontal Component (a)

The horizontal component can be calculated using the formula: \[ a = r \cdot \cos(\theta) \]where \(r\) is the magnitude (which is 3) and \(\theta\) is the angle (which is \(210^{\circ}\)). First, we find \(\cos(210^{\circ})\):\[ \cos(210^{\circ}) = \cos(180^{\circ} + 30^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2} \]Then, calculate \(a\):\[ a = 3 \cdot (-\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} \approx -2.60 \]

Step 3: Calculating the Vertical Component (b)

The vertical component can be calculated using the formula: \[ b = r \cdot \sin(\theta) \]where \(r\) is the magnitude (which is 3), and \(\theta\) is the angle (which is \(210^{\circ}\)). First, we find \(\sin(210^{\circ})\):\[ \sin(210^{\circ}) = \sin(180^{\circ} + 30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2} \]Then, calculate \(b\):\[ b = 3 \cdot (-\frac{1}{2}) = -\frac{3}{2} = -1.50 \]

Step 4: Writing the Vector in Component Form

Combine the calculated horizontal and vertical components to write the vector:The vector can be written as: \[ \mathbf{v} = -2.60i -1.50j \]

Key Concepts

Horizontal Component

When dealing with vectors, the horizontal component is often referred to as the 'x-component'. It represents how far the vector extends along the horizontal axis. To find this component, you need to use a bit of trigonometry. The key formula is:

• The horizontal component, denoted as \(a\), can be calculated using: \(a = r \cdot \cos(\theta)\).

Here, \(r\) represents the magnitude of the vector, while \(\theta\) is the angle the vector makes with the positive x-axis, often referred to as the direction angle. In our example, the magnitude \([r]\) is 3, and the direction angle \([\theta]\) is \(210^{\circ}\).
To find \(\cos(210^{\circ})\), we recognize that this angle is in the third quadrant where cosine values are negative. We rewrite \(210^{\circ}\) as \(180^{\circ} + 30^{\circ}\).

• We know that \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\), so \(\cos(210^{\circ}) = -\frac{\sqrt{3}}{2}\).

Multiply by the magnitude to get the horizontal component:

• \[ a = 3 \times (-\frac{\sqrt{3}}{2}) \approx -2.60 \].

Vertical Component

The vertical component of a vector reflects how much it extends along the vertical axis, also known as the 'y-component'. Calculating this involves the sine function. The formula is:

• Vertical component, \(b = r \cdot \sin(\theta)\).

As with the horizontal component, \(r\) is the magnitude of the vector and \(\theta\) is the direction angle, which is \(210^{\circ}\) in this problem.
In the third quadrant, both sine and cosine are negative. We can find \(\sin(210^{\circ})\) by considering that \(210^{\circ}\) is equivalent to \(180^{\circ} + 30^{\circ}\).

• Since \(\sin(30^{\circ}) = \frac{1}{2}\), it follows that \(\sin(210^{\circ}) = -\frac{1}{2}\).

To find the vertical component:

• \[ b = 3 \times (-\frac{1}{2}) = -1.50 \].

Direction Angle

The direction angle of a vector, often noted as \(\theta\), shows in which direction the vector is pointing. Typically measured from the positive x-axis, it is an essential part of vector analysis in geometry and physics.

• Angles can be in degrees or radians; however, degrees are often easier to visualize.

In this exercise, the direction angle is \(210^{\circ}\), which situates the vector in the third quadrant of a standard coordinate plane. In this quadrant, both x and y coordinates (horizontal and vertical components) are negative.
Understanding this angle's placement helps determine the sign of the components when using trigonometric functions like sine and cosine.

• It's always practical to remember that angles more than \(180^{\circ}\) will fall into either the third or fourth quadrants, where trigonometric functions may have negative values.

Magnitude

The magnitude is the length or size of the vector and is always a positive value, representing the "strength" of the vector. In simpler terms, you can think of it as the distance from the origin of the vector path in a graph.
Magnitude is denoted as \(r\) and in this example, it is given as 3. Combined with the direction angle, the magnitude provides enough information to find both the horizontal and vertical components of the vector.

• You can visualize magnitude as the hypotenuse of a right triangle in vector representation, where the horizontal and vertical components are the other two sides.

Using the Pythagorean theorem visually, for any vector

• \[ \text{Magnitude} = \sqrt{a^2 + b^2} \]

although in exercises like this, it is often already provided.